$\int_0^2 {{e^x}} dx$

Let $I = \int_0^2 {{e^x}} dx$

Here, $a = 0$ and $b = 2$

therefore,$h = \frac{{b - a}}{n}$

$\Rightarrow$ $nh = 2$ and $f(x) = {e^x}$

Now, $\int_0^2 {{e^x}} dx = \mathop {\lim }\limits_{h \to 0} h[f(0) + f(0 + h) + f(0 + 2h) + \ldots + f\{ 0 + (n - 1)h\} ]$

therefore,$I = \mathop {\lim }\limits_{h \to 0} h\left[ {1 + {e^h} + {e^{2h}} + \ldots + {e^{(n - 1)h}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {\frac{{1 \cdot {{\left( {{e^h}} \right)}^n} - 1}}{{{e^h} - 1}}} \right] = \mathop {\lim }\limits_{h \to 0} h\left( {\frac{{{e^{nh}} - 1}}{{{e^h} - 1}}} \right)$

therefore,$I = \mathop {\lim }\limits_{h \to 0} h\left[ {1 + {e^h} + {e^{2h}} + \ldots + {e^{(n - 1)h}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {\frac{{1 \cdot {{\left( {{e^h}} \right)}^n} - 1}}{{{e^h} - 1}}} \right] = \mathop {\lim }\limits_{h \to 0} h\left( {\frac{{{e^{nh}} - 1}}{{{e^h} - 1}}} \right)$

$= \mathop {\lim }\limits_{h \to 0} h\left( {\frac{{{e^2} - 1}}{{{e^h} - 1}}} \right)$

$= {e^2} - 1 = {e^2} - 1$

Evaluate the following :